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Association Structures · Tutorial 172 of 697

Deriving Hazard Ratios for Current-Value Links

Learn to derive and calculate current-value hazard ratios by comparing subjects’ latent marker values on the linear predictor scale.

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What You'll Learn

  • Write the conditional hazard model for a current-value association.
  • Derive the hazard ratio by subtracting two subjects’ log hazards.
  • Identify which terms cancel when baseline covariates are held fixed.
  • Calculate and interpret hazard ratios from marker differences.
  • Recognize when a current-value hazard ratio varies with time.
  • Check assumptions and avoid confusing a conditional hazard ratio with an event-probability ratio.

Why Derive the Contrast?

A current-value association links the event hazard at time \(t\) to a subject’s latent marker value at that same time. To compare two subjects, the relevant quantity is not either hazard in isolation but their ratio at the same time, under the same model. Deriving that ratio makes clear which parts of the model matter for the comparison and how the sign and size of the association coefficient affect the result.

The derivation is conditional on the subjects’ latent trajectories, including their random effects. It therefore describes a subject-specific hazard contrast, not a comparison of event probabilities and not, by itself, a marginal contrast after averaging over random effects. The earlier material on conditional and marginal hazards distinguishes those interpretations.

Model and Assumptions

Let \(h_i(t\mid b_i)\) denote the hazard for subject \(i\) at time \(t\), conditional on the subject’s random effects \(b_i\). Let \(m_i(t)\) be that subject’s latent marker value at time \(t\). In a linear mixed trajectory, this value can be written \(m_i(t)=x_i(t)^\mathsf{T}\beta+z_i(t)^\mathsf{T}b_i\); it is the model-based marker trajectory, rather than a noisy measurement taken at time \(t\).

For a current-value link, write the conditional hazard model as

$$ h_i(t\mid b_i)=h_0(t)\exp\left\{\gamma^\mathsf{T}w_i+\alpha m_i(t)\right\}. $$

Here \(h_0(t)\) is a common baseline hazard, \(w_i\) is a vector of baseline covariates, \(\gamma\) is its coefficient vector, and \(\alpha\) is the current-value association parameter. The model says that, conditional on the random effects and covariates, the log hazard is the sum of a baseline log hazard, a baseline-covariate contribution, and a term proportional to the current latent marker value.

Assumptions for the contrast: Compare two subjects \(A\) and \(B\) at the same time \(t\), when both are in the risk set. The model uses the same positive baseline hazard \(h_0(t)\) and the same coefficients for both subjects. The current-value link is linear on the log-hazard scale, with a common coefficient \(\alpha\). To isolate the marker contrast, hold baseline covariates fixed: \(w_A=w_B\). The comparison is conditional on the latent trajectories.

These assumptions specify the contrast being derived; they are not a claim that every application must use this particular association structure. If covariates differ between the subjects, their contribution must be included. If the link is not linear in \(m_i(t)\), the difference in marker values alone will not generally determine the log hazard ratio in this form.

Deriving the Current-Value Hazard Ratio

Define the conditional hazard ratio for \(A\) relative to \(B\) at time \(t\) as \(h_A(t\mid b_A)/h_B(t\mid b_B)\). Substituting the model for each subject gives

$$ \frac{h_A(t\mid b_A)}{h_B(t\mid b_B)} = \frac{h_0(t)\exp\{\gamma^\mathsf{T}w_A+\alpha m_A(t)\}} {h_0(t)\exp\{\gamma^\mathsf{T}w_B+\alpha m_B(t)\}}. $$

At the same time \(t\), the common baseline hazard cancels. If the baseline covariates are also held fixed, their terms cancel as well. Using the rule \(\exp(u)/\exp(v)=\exp(u-v)\), the result is

$$ \mathrm{HR}_{A:B}(t) = \exp\left\{\alpha\left[m_A(t)-m_B(t)\right]\right\}. $$

Equivalently, the log hazard ratio is the difference between the subjects’ linear predictors:

$$ \log\{\mathrm{HR}_{A:B}(t)\} = \alpha\left[m_A(t)-m_B(t)\right]. $$
Theorem: Under the stated current-value model, for two subjects compared at the same time with identical baseline covariates, the conditional hazard ratio for \(A\) relative to \(B\) is \(\exp\{\alpha[m_A(t)-m_B(t)]\}\). It exceeds 1 when \(\alpha[m_A(t)-m_B(t)]>0\), is below 1 when that product is negative, and equals 1 when the product is zero.

This result is a direct consequence of subtracting the two log hazards. For completeness, taking logs of the model yields \(\log h_i(t\mid b_i)=\log h_0(t)+\gamma^\mathsf{T}w_i+\alpha m_i(t)\). Subtracting the expression for \(B\) from that for \(A\) removes \(\log h_0(t)\); with \(w_A=w_B\), it also removes \(\gamma^\mathsf{T}w_i\). The remaining difference is exactly \(\alpha[m_A(t)-m_B(t)]\). Exponentiating gives the hazard ratio above.

If covariates are not held fixed, the more general contrast is

$$ \mathrm{HR}_{A:B}(t) = \exp\left\{\gamma^\mathsf{T}(w_A-w_B) +\alpha[m_A(t)-m_B(t)]\right\}. $$

Thus, a statement about the marker’s contribution alone requires either matching on baseline covariates or explicitly separating their contribution from the marker contribution.

Worked Linear-Predictor Comparisons

Worked Example: A Two-Unit Marker Difference

Suppose two hypothetical subjects are compared at the same time, have identical baseline covariates, and have latent marker values \(m_A(t)=6\) and \(m_B(t)=4\). Let \(\alpha=0.35\). The common baseline hazard is \(0.02\) per year. Find the conditional hazard ratio for \(A\) relative to \(B\).

The marker difference is \(6-4=2\) marker units. The log hazard ratio is

$$ \alpha[m_A(t)-m_B(t)] =0.35(2)=0.70. $$

Therefore,

$$ \mathrm{HR}_{A:B}(t)=\exp(0.70)=2.014. $$

At this time, subject \(A\)’s conditional hazard is 2.014 times subject \(B\)’s, under the model and conditional on their latent trajectories. The common baseline hazard does not enter the ratio. The marker unit is whatever unit was used to define \(m_i(t)\); \(\alpha\) is expressed per marker unit on the log-hazard scale, so their product is dimensionless.

Worked Example: A Negative Association Coefficient

Suppose \(\alpha=-0.25\), the baseline covariates are identical, and at time \(t=2\) years the latent trajectories give \(m_A(2)=4.0\) and \(m_B(2)=4.4\). Calculate the hazard ratio for \(A\) relative to \(B\).

First find the marker difference, in the order \(A\) minus \(B\): \(4.0-4.4=-0.4\). The log hazard ratio is

$$ (-0.25)(-0.4)=0.10. $$

Exponentiating gives

$$ \mathrm{HR}_{A:B}(2)=\exp(0.10)=1.105. $$

Although \(A\) has the lower marker value, the negative association coefficient means that a lower marker value corresponds to a higher conditional hazard in this model. Thus, at year 2, \(A\)’s conditional hazard is about 1.105 times \(B\)’s. The direction follows from the product of the coefficient and the ordered marker difference, not from the marker difference alone.

Worked Example: Checking the Contrast With Hazards

Consider a hypothetical model with \(h_0(t)=0.02\) per year at the comparison time, a shared baseline-covariate contribution \(\gamma^\mathsf{T}w=0.50\), and \(\alpha=0.20\). Let \(m_A(t)=5\) and \(m_B(t)=3\). Calculate each conditional hazard and then verify the ratio using the derived formula.

The linear predictor apart from the baseline hazard is \(0.50+0.20m_i(t)\). For subject \(A\), it is \(0.50+0.20(5)=1.50\), so

$$ h_A(t\mid b_A)=0.02\exp(1.50) =0.02(4.481689) =0.08963\ \text{per year}. $$

For subject \(B\), it is \(0.50+0.20(3)=1.10\), so

$$ h_B(t\mid b_B)=0.02\exp(1.10) =0.02(3.004166) =0.06008\ \text{per year}. $$

The ratio of these hazards is approximately \(0.08963/0.06008=1.492\). Directly from the current-value contrast,

$$ \exp\{0.20(5-3)\}=\exp(0.40)=1.492. $$

The two calculations agree, up to rounding. The shared covariate contribution affects both hazards but cancels from their ratio.

Time-Specific Comparisons and Diagnostics

The formula is evaluated at a specified time because \(m_A(t)-m_B(t)\) can change as the latent trajectories evolve. Consequently, the conditional hazard ratio need not be constant over follow-up, even though the association coefficient \(\alpha\) is constant. For example, one subject may have a higher latent value early and a lower value later. A single hazard ratio would then conceal how the modeled contrast changes with time.

A useful diagnostic calculation is to write the comparison in three steps: find \(m_A(t)-m_B(t)\), multiply by \(\alpha\), and exponentiate. This makes the ordering and sign visible before rounding. It also helps detect common implementation errors, such as reversing the subject order, using the wrong time point, or treating a noisy observed measurement as though it were the latent current value.

Check the assumptions behind the cancellation. The baseline hazard cancels only because both hazards are evaluated at the same time under a common baseline hazard. A baseline-covariate term cancels only when its contribution is equal for both subjects. If covariates differ, use the general contrast rather than silently dropping the difference. If the model’s association term is not \(\alpha m_i(t)\), this derivation does not apply unchanged.

A frequent mistake is to report \(\alpha[m_A(t)-m_B(t)]\) itself as the hazard ratio. That quantity is the log hazard ratio; the hazard ratio is its exponential. Another mistake is to interpret a hazard ratio of 2 as twice the probability of an event over a follow-up interval. A hazard is an instantaneous event rate among those still at risk, and its ratio is not generally an event-probability ratio.

What the Contrast Does—and Does Not—Compare

This is a conditional, subject-specific comparison: it uses the latent marker values for the two specified subjects. It does not average over the distribution of random effects. After such averaging, a marginal hazard ratio is generally not obtained by simply substituting mean marker values into this formula, and it need not retain the same proportional-hazards form.

The current-value model also differs from approaches that associate the hazard with another feature of the longitudinal trajectory, such as its rate of change, or with a summary over past values. Those approaches answer different scientific questions and have different hazard-ratio derivations. Here the association is specifically with the latent marker level at the time of comparison.

Finally, the formula is a model-based association, not evidence that changing the marker would cause the hazard to change by the corresponding factor. Its interpretation depends on the fitted model, the scale and units of the latent marker, and the validity of the assumptions. The most informative report states the two subjects’ marker values, the comparison time, the value of \(\alpha\), the covariate condition, and which subject is in the numerator.

Check Your Understanding

For each question, use the conditional current-value model and compare subjects at the same time with baseline covariates held fixed. Solutions follow the questions.

  1. If \(\alpha=0.40\), \(m_A(t)=7\), and \(m_B(t)=5\), what is the hazard ratio for \(A\) relative to \(B\)?
  2. If \(\alpha=-0.20\), \(m_A(t)=3\), and \(m_B(t)=6\), which subject has the higher conditional hazard, and by what ratio?
  3. If \(\alpha=0.50\) and \(m_A(t)-m_B(t)=-1\), calculate the hazard ratio for \(A\) relative to \(B\).
  4. If \(\alpha=0.20\) and the marker difference \(m_A(t)-m_B(t)\) remains \(0.5\) at all comparison times, does this model give a time-constant hazard ratio for these subjects? Calculate it.

Solutions. (1) The log hazard ratio is \(0.40(7-5)=0.80\), so \(\mathrm{HR}_{A:B}=\exp(0.80)=2.226\). (2) The log hazard ratio is \((-0.20)(3-6)=0.60\), so \(\mathrm{HR}_{A:B}=\exp(0.60)=1.822\); \(A\) has the higher conditional hazard. (3) The log hazard ratio is \(0.50(-1)=-0.50\), so \(\mathrm{HR}_{A:B}=\exp(-0.50)=0.6065\); \(A\)’s hazard is lower. (4) The log hazard ratio is \(0.20(0.5)=0.10\) at every time, so the ratio is \(\exp(0.10)=1.105\), constant for this pair because their marker difference is constant.

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